Another simple limit problem

Question

L'Hôpital tells us that the following limit is $\frac{\pi}{2}$, but I'm wondering if there is a smart way of evaluating this limit without using L'Hôpital

$$ \lim_{x \to \frac{1}{4}}\frac{\log(\tan(\pi x))}{4x - 1} =?$$

Answer 1

This is just a derivative in disguise:

$$ \frac{\log ( \tan ( \pi x ) )}{4x-1} = \frac{1}{4} \frac{\log ( \tan ( \pi x ) ) - \log ( \tan ( \pi / 4 ) )}{x-1/4}$$

since $\tan(\pi/4) = 1$. So when you do L'Hopital's rule you're really just computing the derivative of the numerator at the point and dividing by 4.

Answer 2

Setting $\displaystyle4x-1=y\implies y\to0, x=\frac{y+1}4=\tan\pi x=\tan\left(\frac\pi4+\frac{\pi y}4 \right)=\frac{1+\tan\frac{\pi y}4}{1-\tan\frac{\pi y}4}$

$$\lim_{x\to\dfrac14}\frac{\ln(\tan(\pi x))}{4x-1}$$

$$=\lim_{y\to0}\frac{\ln\left(1+\tan\frac{\pi y}4\right)}y-\lim_{y\to0}\frac{\ln\left(1-\tan\frac{\pi y}4\right)}y$$

Use $\displaystyle\lim_{h\to0}\frac{\ln(1+h)}h=1$

Answer 3

Completing the solution of the lab, we have using infinites series $$ \frac{1}{1 + h} = 1 - h + h^2 - \ldots \quad |h|<1 \quad \Rightarrow $$ $$ \int\frac{1}{1 + h}dh = \ln(1 + h) + C = h - \frac{h^2}{2} + \frac{h^3}{3} - \ldots $$ For $h = 0 \ \Rightarrow \ C = 0$. Futhermore, $\tan x = x + x^3/3 + \ldots$. Thus, $$ \lim_{y \to 0}\frac{\ln\biggl(1 + \tan \frac{\pi y}{4}\biggr)}{y} = \lim_{y \to 0}\frac{\tan\frac{\pi y}{4} - \frac{1}{2}\tan^2 \frac{\pi y}{4} + \ldots}{y} = \lim_{y \to 0} \frac{\frac{\pi y}{4} + O(y^2)\ldots}{y} = \frac{\pi}{4} $$ Similarly, $$ \lim_{y \to 0}\frac{\ln\biggl(1 - \tan \frac{\pi y}{4}\biggr)}{y} = -\frac{\pi}{4} $$ Thus, $$ \lim_{x \to 1/4}\frac{\ln(\tan(\pi x))}{4x - 1} = \frac{\pi}{2} $$

Answer 4

Using equivalent infinitesimals, we have $$ \lim_{x \to \frac{1}{4}}\frac{\log(\tan(\pi x))}{4x - 1} $$ Let $h=4x-1$, then $$ \lim_{h \to 0}\frac{\log\left(\tan\left(\frac\pi 4 (h+1)\right)\right)}{h} $$ $$= \lim_{h \to 0}\frac{\tan\left(\frac\pi 4 (h+1)\right)-1}{h} $$ $$= 2\lim_{h \to 0}\frac{\tan\left(\frac\pi 4 h\right)}{h\left(1-\tan\left(\frac\pi 4 h\right)\right)}$$ $$= \frac\pi 2\lim_{h \to 0}\frac{h}{h\left(1-\tan\left(\frac\pi 4 h\right)\right)}$$ $$= \frac\pi 2\lim_{h \to 0}\frac{1}{1-\tan\left(\frac\pi 4 h\right)}$$ $$=\frac\pi 2$$